2 Answers
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Here is a coordinate-free version. Let $(V, \psi)$ be a 4-dimensional symplectic space over a field, and let $\omega \in (\wedge^2 V)^{\ast} = \wedge^2(V^{\ast})$ be the nonzero 2-form arising from $\psi$. Then on the 6-dimensional vector space $W = \wedge^2(V)$ the kernel of $\omega$ is a hyperplane $H$ of dimension 5, and on $W$ there is a natural non-degenerate quadratic form $q$ valued in the line $\wedge^4(V)$ via $q(w) = (1/2)(w \wedge w)$ for $w \in W$ (e.g., if $w = e_1 \wedge e_2 + e_3 \wedge e_4$ for a basis $\{e_i\}$ of $W$ then $q(w) = e_1 \wedge e_2 \wedge e_3 \wedge e_4$); the definition of $q$ uses base change from the $\mathbf{Z}_{(2)}$-flat case for $(V,\psi)$ if $2$ isn't a unit.

By computing in linear coordinates of $V$ that "standardize" $\psi$ we see that $q$ is a split quadratic form on $W$, and the action of ${\rm{SL}}(V)$ on $W$ clearly preserves $q$ while the action of its subgroup ${\rm{Sp}}(V,\psi)$ preserves $H$. Hence, the action on $H$ defines a map
$${\rm{Sp}}(V,\psi) \rightarrow {\rm{O}}(q|_H),$$
so this lands inside ${\rm{SO}}(q|_H)$ and as such defines a homomorphism $${\rm{Sp}}_4 = {\rm{Sp}}(V,\psi) \rightarrow {\rm{SO}}(q|_H) = {\rm{SO}}_5.$$

This map kills the center $\mu_2$ inside ${\rm{Sp}}_4$, and thereby identifies ${\rm{Sp}}_4$ as the degree-2 "simply connected" central cover of ${\rm{SO}}_5$ (in the sense of algebraic groups). Hence, this uniquely lifts to an isomorphism of ${\rm{Sp}}_4$ onto ${\rm{Spin}}_5$.

A nice feature of this conceptual construction is that it kills two birds with one stone: if we don't restrict to $H$ and instead work with the entire 6-dimensional $W$ then a similar construction defines the isomorphism of ${\rm{SL}}_4 = {\rm{SL}}(V)$ onto ${\rm{Spin}}_6$.

To avoid confusion, I think you should perhaps write $\operatorname{SO}_{3,2}$ and $\operatorname{Spin}_{3,2}$ for $\operatorname{SO}_5$ and $\operatorname{Spin}_5$, respectively, and similarly, $\operatorname{Spin}_{3,3}$ for $\operatorname{Spin}_6$ in the last line.
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José Figueroa-O'FarrillJul 15 '13 at 10:20

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@Jose: I am using standard notation among algebraists who study algebraic groups over an arbitrary field $k$, where non-degenerate quadratic forms are not classified by a "signature", but typically a lot more data. One writes ${\rm{SO}}(q)$ and ${\rm{Spin}}(q)$ for the connected semisimple $k$-groups associated to a non-degenerate quadratic space $(V,q)$, and when $q$ is split with $V$ of rank $n$ then they are denoted ${\rm{SO}}_n$ and ${\rm{Spin}}_n$. For $k=\mathbf{R}$, one writes ${\rm{SO}}(r,s)$ to denote ${\rm{SO}}(q)(k)$ for $q$ of signature $(r,s)$. So I prefer to leave it as is.
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user36938Jul 15 '13 at 12:36

I do agree it's good to learn how to avoid "choice-of-coordinates" silliness. But/and some of the intrinsification is better understood after one has done things an ad-hoc way. (I think we all know this, although it is not high-rep to admit it.) Then there is the field-specific issue of "classification" of (non-degenerate) quadratic forms over a particular field, which is surely not solvable in general... But, over any algebraically closed field of char not 2 (such as complex...) it's just dimension, and a slightly non-trivial result: over $\mathbb R$, it's "signature". Not "just algebra".
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paul garrettJul 15 '13 at 23:55

@paul: even in char. 2 everything works perfectly well based on char-free definitions, as it must if there is to be a reasonable theory over $\mathbf{Z}$ for linking it up with Chevalley groups. The "correct" dichotomy for SO and Spin groups is not char. 2 versus char. $\ne 2$ but rather $\dim V$ being even or odd (i.e., B versus D).
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user36938Jul 16 '13 at 1:19

Edit: Over algebraically closed fields (especially in char not $2$, which I want to not think about), these sporadic isogenies are easy to write down. In principle, but I think not in practice for most of us, to see what happens over not algebraically closed fields is a question of "Galois cohomology", as in Weil's "Algebras and classical groups" paper. For me, it's much easier to use a few coordinates. E.g., although $SU(4)\rightarrow SO(6)$ and $SU(2,2)\rightarrow SO(4,2)$, apparently $SU(3,1)$ does not map to $SO(p,q)$ with $p+q=6$. Meanwhile, $SL_2(\mathbb H)\rightarrow SO(5,1)$. Seems weird to me.

For completeness: Paul lets $Sp_4(R)$ act on the 16-dimensional matrix algebra $M_4(R)$ by conjugation. The usual bilinear form $tr(xy)$ is invariant. The Lie algebra $RI_4+sp_4(R)$ is an invariant subspace on which this form is nondegenerate ($sp_4(R)$ being the 10-dimensional Lie algebra), so its orthogonal is also nondegenerate, it is thus (16-(10+1))-dimensional, i.e. 5-dimensional. Explicit coordinates show the signature is (3,2). This defines a 2-to-1 map from $Sp_4(R)$ onto $SO(3,2)$.
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YCorJul 14 '13 at 21:57

@paulgarrett: $SU(3,1)$ that you ask about maps to $SO^*(6)$ (a.k.a. $SO(3,\mathbb H)$), according to Helgason (§X.6.4) or Besse (Einstein Manifolds, p. 201). You may need to add a third section "Over $\mathbb H$"!
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Francois ZieglerJul 23 '13 at 4:30